# Thread: Determining straight line characteristics for 1st order PDE's

1. ## Determining straight line characteristics for 1st order PDE's

Dear Folks,

Calculate the straight line characteristics for the following linear 1st order PDE's

1) $\displaystyle 3u_x +5u_y -xyu=0$

I calculate $\displaystyle k=y-\frac{5}{3}x$ and when transformed becomes $\displaystyle \eta=y-\frac{5}{3}\xi$

but the book its given as $\displaystyle 5x-3y=k$ and when transformed becomes $\displaystyle 5\xi-3y=\eta$

2) $\displaystyle yu_x+x^2uy=xy$

I calculate $\displaystyle k=-\frac{x^3}{3}+\frac{y^2}{2}$ when transformed becomes $\displaystyle \eta=\frac{-\xi^3}{3}+\frac{y^2}2}$

but the book is $\displaystyle k=2x^3-3y^2$ when transformed becomes $\displaystyle \eta=2\xi^3-3y^2$
The only way I can see them arriving at this is $\displaystyle k'=\frac{-x^3}{3}+\frac{y^2}{2}$ therefore
$\displaystyle -6k'=2x^3-3y^2=k$

The pattern I see in these kind of questions is that when k is written in terms of x and y that the 'x' terms are always positve before transforming. Why is this?

Thanks

2. It all comes down to a matter of prefrence. The writter of the solutions (I addition to myself) doesn't like fractions and prefers x before y.

3. Originally Posted by Danny
It all comes down to a matter of prefrence. The writter of the solutions (I addition to myself) doesn't like fractions and prefers x before y.
so its ok to change the sign so long as its accounted for in the constant -6k'=k etc as above? Ie, $\displaystyle k=-\frac{x^3}{3}+\frac{y^2}{2}$ becomes $\displaystyle k=2x^3-3y^2$

Thanks

4. Let me show you by example.

$\displaystyle y u_x + x^2 u_y = xy$

Let me choose

(1) $\displaystyle r = - \dfrac{x^3}{3} + \dfrac{y^2}{2},\;\;\; s = x$

(2) $\displaystyle r = \dfrac{x^3}{3} - \dfrac{y^2}{2},\;\;\; s = x$

(3) $\displaystyle r = 2x^3 - 3y^2,\;\;\; s = x$.

All three transform your PDE to

$\displaystyle u_s = s\;\;\; \text{so}\;\;\; u = \dfrac{s^2}{2} + F(r)$.

(1) $\displaystyle u = \dfrac{x^2}{2} + F\left( - \dfrac{x^3}{3} + \dfrac{y^2}{2}\right)$

(2) $\displaystyle u = \dfrac{x^2}{2} + F\left( \dfrac{x^3}{3} - \dfrac{y^2}{2}\right)$

(3) $\displaystyle u = \dfrac{x^2}{2} + F\left(2x^3 - 3y^2 \right)$.

As F is arbitrary you can always choose $\displaystyle F(\lambda) = F(k \cdot \lambda),$ thus making all three equivalent.

5. As F is arbitrary you can always choose $\displaystyle F(\lambda) = F(k \cdot \lambda),$ thus making all three equivalent.
so $\displaystyle F(-6(\frac{-x^3}{3}+\frac{y^2}{2}))=F(2x^3-3y^2)$ where $\displaystyle \lambda=-6$

Im happy with that. Thanks Danny

6. sorry, that should be k=-6
cheers

7. What I really should have said to be consistent with your notation is

$\displaystyle F(k) = F(\lambda k)$.